Question

The formula that relates all three elastic constants is.
A. $9/y = 3/n - 1/B$
B. $9/y = 3/n + 1/B$
C. $9/y = 3/n + 2/B$
D. $9/y = 3/n - 2/B$

Answer 1

Hint: To build the relation between the three elastic constants, we can use the relations among two elastic constants with each other. To determine the answer in the form of given options, we can do some simple rearrangement and calculation with the relations of elastic constants.

Complete step by step solution:
Let $y$ is the modulus of elasticity, $n$ is the modulus of rigidity and $B$ is the bulk modulus.
The relation between $y$ and $n$ is,
$y = 2n\left( {1 + \mu } \right)$ (1)

Here $\mu $ is the Poisson’s ratio.

The relation between $y$ and $B$ is,
$y = 3B\left( {1 - 2\mu } \right)$ (2)

From equation (2), the expression of the Poisson’s ratio is,
$\begin{array}{l}
y = 3B\left( {1 - 2\mu } \right)\\
\dfrac{y}{{3B}} - 1 = - 2\mu \\
\mu = \dfrac{1}{2} - \dfrac{y}{{6B}}
\end{array}$ (3)

Now from equation (1) and (3), the relation between all three elastic constant is,
\[\begin{array}{l}
y = 2n\left( {1 + \left( {\dfrac{1}{2} - \dfrac{y}{{6B}}} \right)} \right)\\
y = 2n + n - \dfrac{{ny}}{{3B}}\\
y = \dfrac{{9nB - ny}}{{3B}}\\
3yB + nY = 9nB
\end{array}\]

On further solving the above equation
\[\begin{array}{l}
y = \dfrac{{9nB}}{{3B + n}}\\
\dfrac{9}{y} = \dfrac{3}{n} + \dfrac{1}{B}
\end{array}\]

Therefore, the option (B) is the correct answer that is $9/y = 3/n + 1/B$.

Note: The relation of elastic constants consist of the value of Poisson's ratio, so do not forget to remove the value of Poisson's ratio from the relations. To remove Poisson's ratio, replace the value of Poisson's ratio arrived from equation (2) in equation (1).